My pde lecture defined
$W^{1,p}_0(a,b):=clos_{\lVert \cdot \rVert_{1,p}} \mathcal{C}^\infty_c(a,b)$
For $p<\infty$ we proved that
$W^{1,p}_0(a,b)=\{u \in W^{1,p}(a,b), u(a)=0=u(b)\}$
where this pointwise evaluation is to be understood as the pointwise evalution of the continuous representative.
The proof doesn`t work for $p=\infty$ and in fact we were told, that the statement is not not true for this choice of p. (Though the $\subset$ inclusion should still hold, right?).
Can somebody provide an example which is not in the closure on the left?
Also, I was wondering if there is any intuition behind $W^{1, \infty}_0(a,b)$. How does this space look and how is it useful?
For $q>1$ we also defined
$W^{-1,q}(a,b):=(W^{1,p}_0(a,b))^*$
The notation makes sense, since $W_0^{1,p}(a,b) \overset{\mathrm{d}}{\hookrightarrow} L^p(a,b)$ (compactly for $p>1$) and $L^q(a,b)\cong (L^p(a,b))^*$ for $p<\infty$.
Again, I was wondering about the case of $p=\infty$ or $q=1$. Is $W^{-1,1}(a,b)$ even defined? Is there any use or intuation behind this space?
(The notation clearly woudn´t make a lot of sense, since $L^1(a,b) \not\cong (L^\infty(a,b))^*$, but we can still think about the dual space on the right hand side).
I also have a question about the hilbert space case; I was wondering about the folowing statement in my script:
For all $f \in H^{-1}(a,b)(:=W^{-1,2}(a,b))$ there exists a (not unique) $v_f \in L^2(a,b)$ s.t. $$\forall \ u \in H^1(a,b)(:=W^{1,2}(a,b)): \langle f, u \rangle=f(u)=\int_a^b v_f(x)u^\prime(x) \mathrm{d}x=(v_f,u)_{1,2}$$
Now, what puzzles me about this statement is the "not unique" part. For $f \in (L^2(a,b))^*$ the existence of a unique $v_f \in L^2(a,b)$ with $f(u)=(v_f, u)_{1,2}$ for all $u \in L^2(a,b)$ is implied by Riez`s representation theorem. But is $f$ not already completely characterized by $f|_{W^{1,2}_0(a,b)}$, since this space is dense in $L^2(a,b)$? Shoudn´t the $u_f$ in the theorem from my script be unique?
Thanks in advance for any answers as well as corrections of what I wrote :)
You know, classically, that $C_0(X),$ where $X$ is locally compact, is a closed subspace of $L^{\infty}$. The exact same argument tells you that $C_0^k(X)$ is closed in the norm $$ ||f||_{k,\infty}=\sum_{j=0}^k \sum_{|\alpha|=j}||\partial^{\alpha} f||_{\infty}, $$ and this is exactly the $W^{k,\infty}$-norm. Hence, to answer your second question: $W^{k,\infty}$ is always a classical function space.
To answer your first question, if you take $f(x)=1_{[0,1)}-1_{[1,2]},$ then $g(x)=\int_0^{x} f(t)\textrm{d}t$ defines an element of $L^{\infty}_{loc}(\mathbb{R})$ and hence, it restricts to an $L^{\infty}$ function on any bounded interval. It is also immediate, that $g(0)=0$ and $g(2)=0,$ but $g$ does not have a continuous derivative, hence does not belong to the space above with $k=1$ and $X=(0,2)$.